Question

Question 3 [25 points]: Complex integration Subquestions (a), (b), and (c) will use C1 shown in the figure on the left-hand side, whereas subquestion (d) will use C2 shown in the figure on the right-hand side. Im (2) Im (2) SA= 1 → Re (2) → Re (2) 20 = 1 - (a) [3 points) Find a parametric representation for the curve Ci. (b) [7 points] Compute the integral Sc, z dz. (c) [5 points) Compute the integral Se, 22 dz. (d) [10 points] Consider the curve C2 with the parameterization z(t) 4 sin(2t) +i.sin(t), A<t< 27. Compute the integral be, ztdz. Justify your calculations!

Answer 1

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quarker As as ist part of circle, centered at toate & radius 1. - 17-01-21 = 1 70=1-2, This z(t) = x(t) + 1 y(t), where act) = 1 + cos (1/2+t). ç yct)= -1 + sin(974 ++) for t€ (0,72] is parametrization of C curve. We can check 20) = 200) + 1 400) = 1 8 7(1) = -l are starting & ending points of C. 6 To compute s į dz As Z= a(t) tiyct) dz= -sin (7/274) + è cos (76+t) dt This 1787 = 5 12+ cos +t) - i (-1 +sin fiy+e))][sin(4+4)# icoft te dt T. pois a potenti di + Si el reto), 1700 i (trett) Lo Here we are cesing z(t) =(-i)+€ =(-1) ) dt + sidt + i Tyq = (-1){ [41] to the = -210-é) te te S zdz =-2+(2+4 / ++

To compute z dz Let I(Z) = 7² then g(z) = 2 is primitive of function f(that is gl=f.) Thus (zedz has value independent of the contour Its value is determined by endpoints of C, by 9 (2) - Thus fadz = 9 (73)-3(70) 3 Post Z -- or I E be curve with parametrization Z(t) = sinist tisint, tist<27 To compute 6 1 dz -d i dz 7- / ) (2+1/2) 741 - dz. & z-vzi de 9 (7+vi) Z-(hi) Now do I (Z) = 1 is analytic except xe but yi& closed path As Yi does not lies inside of Co, which is closed Now the dit) - zob Z=12 l

by cauchy integral e formula & grey dz = 271 g(zo) Z-Zo Thus 6 ?dz = 212 9(70) = 271 Lumi = 21 7²+1/4 de = 251